# Thread: Logarithms and factorial; identity

1. ## Logarithms and factorial; identity

Hi Forum
Can someone help?
There's a question that states

$1+\log_{n!} {(n+1)} = \log_{n!} {(n+1)!}$

But how was this thought?

Where could I find about this identity on the web?

The question is actually
$\displaystyle\prod_{n=2}^{10} (1+\log_{n!} {(n+1)})$

Does it change anything by being Cartesian product?
Thanks

2. ## Re: Logarithms and factorial; identity

Hello, Zellator!

$1+\log_{n!}(n+1) \:=\: \log_{n!}(n+1)!$

$\text{But how was this thought?}$

Note that: . $\log_{n!}(n!) \:=\:1$

The left side becomes:
. . $\log_{n!}(n!) + \log_{n!}(n+1) \;=\;\log_{n!}\bigg[n!(n+1)\bigg] \;=\;\log_{n!}(n+1)!$

3. ## Re: Logarithms and factorial; identity

Originally Posted by Zellator
Hi Forum
Can someone help?
There's a question that states

$1+\log_{n!} {(n+1)} = \log_{n!} {(n+1)!}$

But how was this thought?

Where could I find about this identity on the web?

The question is actually
$\displaystyle\prod_{n=2}^{10} (1+\log_{n!} {(n+1)})$

Does it change anything by being Cartesian product?
Thanks
Working from the right

$log_{n!}(n+1)!=log_{n!}\left[(n+1)n!\right]=log_{n!}(n!)+log_{n!}(n+1)$

4. ## Re: Logarithms and factorial; identity

Originally Posted by Soroban
Hello, Zellator!

Note that: . $\log_{n!}(n!) \:=\:1$

The left side becomes:
. . $\log_{n!}(n!) + \log_{n!}(n+1) \;=\;\log_{n!}\bigg[n!(n+1)\bigg] \;=\;\log_{n!}(n+1)!$

Then using $n!(n+1)$ we can see that $(n+1)(n)(n-1)!$
So $n!(n+1)= (n+1)!$

But that's kind of hard to see in the naked eye, thanks to you I will probably find it this forms easily from now on.

Originally Posted by Archie Meade
Working from the right

$log_{n!}(n+1)!=log_{n!}\left[(n+1)n!\right]=log_{n!}(n!)+log_{n!}(n+1)$
Hi Archie Meade,
The thing is, that the right form wasn't given, it had to be calculated, in order to find a better and easier way to solve the Cartesian Product.

5. ## Re: Logarithms and factorial; identity

Ok,

well that's important to realise about factorials.

$6!=(6)5!$

$7!=(7)6!$

and so on.

Also, a very fundamental property of logarithms is

$log_{a}a=y\Rightarrow\ a=a^y\Rightarrow\ y=1$

Then

$1+log_{n!}(n+1)=1+log_{n!}\frac{n!(n+1)}{n!}=1+log _{n!}\frac{(n+1)!}{n!}$

$=1+log_{n!}(n+1)!-log_{n!}(n!)$